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Q.
If $f(x)=\pi\left(\frac{\sqrt{x+7}-4}{x-9}\right)$, then the range of function $y=\sin (2 f(x))$ is
Relations and Functions - Part 2
Solution:
funcisc Given, $f(x)=\pi\left(\frac{\sqrt{x+7}-4}{x-9}\right)$
Clearly, domain of $f(x)=[-7, \infty)-\{9\}$.
Now, $f(x)=\frac{\pi(x+7-16)}{(x-9)(\sqrt{x+7}+4)}$ (Rationalise $)=\frac{\pi}{\sqrt{x+7}+4}$.
So, range of $f(x)$ is $\left(0, \frac{\pi}{4}\right]-\left\{\frac{\pi}{8}\right\}$.
Hence, range of $y=\sin (2 f(x))$ is $(0,1]-\left\{\frac{1}{\sqrt{2}}\right\}$. Ans.]